Points (6 ,7 ) and (7 ,5 ) are (2 pi)/3 radians apart on a circle. What is the shortest arc length between the points?

1 Answer
Feb 17, 2016

2/9*sqrt(15)*pi~=2.704

Explanation:

Refer to the figure below

I created this figure using MS ExcelI created this figure using MS Excel

alpha=(2pi)/3 radians
Or
alpha=(2*180^@)/3=120^@

AB=sqrt((7-6)^2+(5-7)^2)=sqrt(1+4)=sqrt(5)

Applying Law of Cosines in triangle_(ABC):

AB^2=r^2+r^2-2r*r*cos 180^@
2r^2-2r^2*(-1/2)=(sqrt(5))^2
3r^2=5 => r=sqrt(5/3)

Length of the arc

arc AB=r*alpha, where alpha is given in radians
arc AB=sqrt(5/3)*(2*pi)/3*(sqrt(3)/sqrt(3))=2/9*sqrt(15)*pi